Matrices

We know that in mathematics, a matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

The number of rows and columns that a matrix has is called its order or its dimension. By convention, rows are listed first; and columns second.

We are given with a matrix that has 28 elements.

We know that,

If a matrix has mn elements, then the order of the matrix can be given by m × n, where m and n are natural numbers.

Therefore, for a matrix having 28 elements, that is, mn = 28, possible orders can be found out as follows:

∵ mn = 28

Take m and n to be any number, such that, when it is multiplied it gives 28.

So, let m = 1 and n = 28.

Then, m × n = 1 × 28 (=28)

⇒ 1 × 28 is a possible order of the matrix having 28 elements.

Take m = 2 and n = 14.

Then, m × n = 2 × 14 (=28)

⇒ 2 × 14 is a possible order of the matrix having 28 elements.

Take m = 4 and n = 7.

Then, m × n = 4 × 7 (=28)

⇒ 4 × 7 is a possible order of the matrix having 28 elements.

Take m = 7 and n = 4.

Then, m × n = 7 × 4 (=28)

⇒ 7 × 4 is a possible order of the matrix having 28 elements.

Take m = 14 and n = 2.

Then, m × n = 14 × 2 (=28)

⇒ 14 × 2 is a possible order of the matrix having 28 elements.

Take m = 28 and n = 1.

Then, m × n = 28 × 1 (=28)

⇒ 28 × 1 is a possible order of the matrix having 28 elements.

Thus, the possible orders of the matrix having 28 elements are

1 × 28, 2 × 14, 4 × 7, 7 × 4, 14 × 2 and 28 × 1

If the matrix had 13 elements, then also we find the possible order in same way.

Here, mn = 13.

Take m and n to be any number, such that, when it is multiplied it gives 13.

Take m = 1 and n = 13.

Then, m × n = 1 × 13 (=13)

⇒ 1 × 13 is a possible order of the matrix having 13 elements.

Take m = 13 and n = 1.

Then, m × n = 13 × 1 (=13)

⇒ 13 × 1 is a possible order of the matrix having 13 elements.

Thus, the possible orders of the matrix having 13 elements are

1 × 13 and 13 × 1

We have the matrix

A matrix, as we know, is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

(i). We need to find the order of the matrix A.

And we know that,

The number of rows and columns that a matrix has is called its order or its dimension. By convention, rows are listed first; and columns second.

So,

Here, in matrix A:

There are 3 rows.

Elements in 1^st row = a, 1, x

Elements in 2^nd row = 2, √3, x² – y

Elements in 3^rd row = 0, 5, -2/5

⇒ M = 3

And,

There are 3 columns.

Elements in 1^st column = a, 2, 0

Elements in 2^nd column = 1, √3, 5

Elements in 3^rd column = x, x² – y, -2/5

⇒ N = 3

Since, the order of matrix = M × N

⇒ The order of matrix A = 3 × 3

Thus, the order of the matrix A is 3 × 3.

(ii). We need to find the number of elements in the matrix A.

And we know that,

Each number that makes up a matrix is called an element of the matrix.

So,

If a matrix has M rows and N columns, the number of elements is MN.

Here, in matrix A:

There are 3 rows.

⇒ M = 3

And,

There are 3 columns.

⇒ N = 3

Then, number of elements = MN

⇒ Number of elements = 3 × 3

⇒ Number of elements = 9

The elements are namely, a, 2, 0, 1, √3, 5, x, x² – y, -2/5.

Thus, the number of elements is 9.

(iii). We need to find the elements a₂₃, a₃₁ and a₁₂.

We know that,

a_ij is the representation of elements lying in the i^th row and j^th column.

For a₂₃:

Comparing a_ij with a₂₃, we have

i = 2

j = 3

Look up in matrix A for element in (i=) 2^nd row and (j=) 3^rd column.

Element that is common to both 2^nd row and 3^rd column = x² – y

⇒ a₂₃ = x² – y

For a₃₁:

Comparing a_ij with a₃₁, we have

i = 3

j = 1

Look up in matrix A for element in (i=) 3^rd row and (j=) 1^st column.

Element that is common to both 3^rd row and 1^st column = 0

⇒ a₃₁ = 0

For a₁₂:

Comparing a_ij with a₁₂, we have

i = 1

j = 2

Look up in matrix A for element in (i=) 1^st row and (j=) 2^nd column.

Element that is common to both 1^st row and 2^nd column = 1

⇒ a₁₂ = 1

Thus, a₂₃ = x² – y, a₃₁ = 0 and a₁₂ = 1.

We know that,

A matrix, as we know, is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

Also,

We know that, the notation A = [a_ij]_m×m indicates that A is a matrix of order m × n, also 1 ≤ i ≤ m, 1 ≤ j ≤ n; i, j ∈ N.

(i).We need to construct a matrix, a_2×2, where

For a_2×2,

1 ≤ i ≤ m

⇒ 1 ≤ i ≤ 2 [∵ m = 2]

And,

1 ≤ j ≤ n

⇒ 1 ≤ j ≤ 2 [∵ n = 2]

Put i = 1 and j = 1.

Put i = 1 and j = 2.

Put i = 2 and j = 1.

⇒ a₂₁ = 0

Put i = 2 and j = 2.

⇒ a₂₂ = 2

Let the matrix formed be A.

Substituting the values of a₁₁, a₁₂, a₂₁ and a₂₂, we get the matrix

(ii). We need to construct a matrix, a_2×2, where

a_ij = |-2i + 3j|

For a_2×2,

1 ≤ i ≤ m

⇒ 1 ≤ i ≤ 2 [∵ m = 2]

And,

1 ≤ j ≤ n

⇒ 1 ≤ j ≤ 2 [∵ n = 2]

Put i = 1 and j = 1.

a₁₁ = |-2(1) + 3(1)|

⇒ a₁₁ = |-2 + 3|

⇒ a₁₁ = |1|

⇒ a₁₁ = 1

Put i = 1 and j = 2.

a₁₂ = |-2(1) + 3(2)|

⇒ a₁₂ = |-2 + 6|

⇒ a₁₂ = |4|

⇒ a₁₂ = 4

Put i = 2 and j = 1.

a₂₁ = |-2(2) + 3(1)|

⇒ a₂₁ = |-4 + 3|

⇒ a₂₁ = |-1|

⇒ a₂₁ = 1

Put i = 2 and j = 2.

a₂₂ = |-2(2) + 3(2)|

⇒ a₂₂ = |-4 + 6|

⇒ a₂₂ = |2|

⇒ a₂₂ = 2

Let the matrix formed be A.

Substituting the values of a₁₁, a₁₂, a₂₁ and a₂₂, we get the matrix

A matrix, as we know, is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

Also,

We know that, the notation A = [a_ij]_m×m indicates that A is a matrix of order m × n, also 1 ≤ i ≤ m, 1 ≤ j ≤ n; i, j ∈ N.

We need to construct a 3 × 2 matrix whose elements are given by

a_ij = e^i.x sin jx

For a_3×2:

1 ≤ i ≤ m

⇒ 1 ≤ i ≤ 3 [∵ m = 3]

1 ≤ j ≤ n

⇒ 1 ≤ j ≤ 2 [∵ n = 2]

Put i = 1 and j = 1.

a₁₁ = e^(1)x sin (1)x

⇒ a₁₁ = e^x sin x

Put i = 1 and j = 2.

a₁₂ = e^(1)x sin (2)x

⇒ a₁₂ = e^x sin 2x

Put i = 2 and j = 1.

a₂₁ = e^(2)x sin (1)x

⇒ a₂₁ = e^2xsin x

Put i = 2 and j = 2.

a₂₂ = e^(2)x sin (2)x

⇒ a₂₂ = e^2x sin 2x

For i = 3 and j = 1.

a₃₁ = e^(3)x sin (1)x

⇒ a₃₁ = e^3x sin x

For i = 3 and j = 2.

a₃₂ = e^(3)x sin (2)x

⇒ a₃₂ = e^3x sin 2x

Let the matrix formed be A.

Substituting the values of a₁₁, a_12, a₂₁, a₂₂, a₃₁ and a₃₂, we get the matrix

Thus, we have got the matrix.

We have the matrices A and B, where

We need to find the values of a and b.

We know that, if

Then,

a₁₁ = b₁₁

a₁₂ = b₁₂

a₂₁ = b₂₁

a₂₂ = b₂₂

Also, A = B.

This means,

a + 4 = 2a + 2 …(i)

3b = b² + 2 …(ii)

8 = 8

-6 = b² – 5b …(iii)

From equation (i), we can find the value of a.

a + 4 = 2a + 2

⇒ 2a – a = 4 – 2

⇒ a = 2

From equation (ii), we can find the value of b².

3b = b² + 2

⇒ b²= 3b – 2

Substitute the value of b² in equation (iii), we get

-6 = b² – 5b

⇒ -6 = (3b – 2) – 5b

⇒ -6 = 3b – 2 – 5b

⇒ -6 = 3b – 5b – 2

⇒ -6 = -2b – 2

⇒ 2b = 6 – 2

⇒ 2b = 4

⇒ b = 2

Thus, a = 2 and b = 2.

We know that,

The number of rows and columns that a matrix has is called its order or its dimension. By convention, rows are listed first; and columns second.

Also,

Addition or subtraction of matrices is possible only if the matrices are of same order.

That is,

If A and B are two matrices and if they are needed to be added, then if order of A is m × n, order of B must be m × n.

We have matrices A and B, where

We know what order of matrix is,

If a matrix has M rows and N columns, the order of matrix is M × N.

In matrix A:

Number of rows = 2

⇒ M = 2

Number of column = 2

⇒ N = 2

Then, order of matrix A = M × N

⇒ Order of matrix A = 2 × 2

In matrix B:

Number of rows = 2

⇒ M = 2

Number of columns = 3

⇒ M = 3

Then, order of matrix B = M × N

⇒ order of matrix B = 2 × 3

Since,

Order of matrix A ≠ Order of matrix B

⇒ Matrices A and B cannot be added.

Thus, matrix A and matrix B cannot be added.

Addition or subtraction of matrices is possible only if the matrices are of same order.

That is,

If A and B are two matrices and if they are needed to be added, then if order of A is m × n, order of B must be m × n.

We have matrices X and Y, where

We know what order of matrix is,

If a matrix has M rows and N columns, the order of matrix is M × N.

(i). We need to find the X + Y.

Let us first determine order of X and Y.

Order of X:

Number of rows = 2

⇒ M = 2

Number of columns = 3

⇒ N = 3

Then, order of matrix X = M × N

⇒ Order of matrix X = 2 × 3

Order of Y:

Number of rows = 2

⇒ M = 2

Number of columns = 3

⇒ N = 3

Then, order of matrix Y = M × N

⇒ Order of matrix Y = 2 × 3

Since, order of matrix X = order of matrix Y

⇒ Matrices X and Y can be added.

So,

Thus, .

(ii). We need to find 2X – 3Y.

Let us calculate 2X.

We have,

Then, multiplying by 2 on both sides, we get

Also,

Multiplying by 3 on both sides, we get

Now subtract 3Y from 2X.

Thus, .

(iii). We need to find matrix Z, such that X + Y + Z is a zero matrix.

That is,

X + Y + Z = 0

Or,

Z = -X – Y

Or,

Z = -(X + Y)

We have already found X + Y in part (i).

So, from part (i):

Then,

Thus, .

A matrix, as we know, is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

Also,

Two or more matrices can be added or subtracted only if they have same order.

And we are familiar with order of a matrix. If a matrix has M rows and N columns, the order of matrix is M × N.

We have matrix equation,

Take matrix .

Multiply it with x,

…(i)

Take matrix .

Multiply it with 2,

…(ii)

Take matrix .

Multiply it with 2,

…(iii)

Add equation (i) and (ii) and make it equal to equation (iii), we get

Adding left side of the matrix equation as they have same order.

We need to find the value of x.

So, compare the elements in the two matrices.

If,

Then,

a₁₁ = b₁₁

a₁₂ = b₁₂

a₂₁ = b₂₁

a₂₂ = b₂₂

So,

2x² + 16 = 2(x² + 8) …(i)

2x + 10x = 48 …(ii)

3x + 8 = 20 …(iii)

x² + 8x = 12x …(iv)

We have got equations (i), (ii), (iii) and (iv) to solve for x.

So, take equation (i).

2x² + 16 = 2x² + 16

We won’t be able to find x from this equation, as both equations are same.

Now, take equation (ii).

2x + 10x = 48

⇒ 12x = 48

⇒ x = 4

From equation (iii),

3x + 8 = 20

⇒ 3x = 20 – 8

⇒ 3x = 12

⇒ x = 4

From equation (iv),

x² + 8x = 12x

⇒ x² = 12x – 8x

⇒ x² = 4x

⇒ x² – 4x = 0

⇒ x(x – 4) = 0

⇒ x = 0 or (x – 4) = 0

⇒ x = 0 or x = 4

⇒ x = 4 (∵ x = 0 does not satisfy equations (ii) and (iii))

So, by solving equations (ii), (iii) and (iv), we can conclude that

x = 4

Thus, the value of x is 4.

We have the matrices A and B, where

We need to show that (A + B) (A – B) ≠ A² – B².

Take L.H.S: (A + B) (A – B)

First, let us compute (A + B).

If two matrices are of same order (say, m × n), then they can be added or subtracted. Example,

If we have matrices and . Then, they can be added as

So,

Now, let us compute (A – B).

Similarly, two matrices having same order can be subtracted in a similar fashion.

So,

Now, let us compute (A + B) (A – B).

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(0, 0).(0, 0) = (0 × 0) + (0 × 0)

⇒ (0, 0).(0, 0) = 0 + 0

⇒ (0, 0).(0, 0) = 0

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(0, 0).(2, 1) = (0 × 2) + (0 × 1)

⇒ (0, 0).(2, 1) = 0 + 0

⇒ (0, 0).(2, 1) = 0

Multiply 2^nd row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(2, 1).(0, 0) = (2 × 0) + (1 × 0)

⇒ (2, 1).(0, 0) = 0 + 0

⇒ (2, 1).(0, 0) = 0

Multiply 2^nd row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(2, 1).(2, 1) = (2 × 2) + (1 × 1)

⇒ (2, 1).(2, 1) = 4 + 1

⇒ (2, 1).(2, 1) = 5

So, we have

Take R.H.S: A² – B²

Let us compute A² first.

A² = A.A

So, we need to compute A.A.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix A, then sum them up.

(0, 1).(0, 1) = (0 × 0) + (1 × 1)

⇒ (0, 1).(0, 1) = 0 + 1

⇒ (0, 1).(0, 1) = 1

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix A, then sum them up.

(0, 1).(1, 1) = (0 × 1) + (1 × 1)

⇒ (0, 1).(1, 1) = 0 + 1

⇒ (0, 1).(1, 1) = 1

Multiply 2^nd row of matrix A by matching members of 1^st column of matrix A, then sum them up.

(1, 1).(0, 1) = (1 × 0) + (1 × 1)

⇒ (1, 1).(0, 1) = 0 + 1

⇒ (1, 1).(0, 1) = 1

Multiply 2^nd row of matrix A by matching members of 2^nd column of matrix A, then sum them up.

(1, 1).(1, 1) = (1 × 1) + (1 × 1)

⇒ (1, 1).(1, 1) = 1 + 1

⇒ (1, 1).(1, 1) = 2

So,

Now, let us compute B².

B² = B.B

We need to compute B.B.

Multiply 1^st row of matrix B by matching members of 1^st column of matrix B, then sum them up.

(0, -1).(0, 1) = (0 × 0) + (-1 × 1)

⇒ (0, -1).(0, 1) = 0 – 1

⇒ (0, -1).(0, 1) = -1

Multiply 1^st row of matrix B by matching members of 2^nd column of matrix B, then sum them up.

(0, -1).(-1, 0) = (0 × -1) + (-1 × 0)

⇒ (0, -1).(-1, 0) = 0 + 0

⇒ (0, -1).(-1, 0) = 0

Multiply 2^nd row of matrix B by matching members of 1^st column of matrix B, then sum them up.

(1, 0).(0, 1) = (1 × 0) + (0 × 1)

⇒ (1, 0).(0, 1) = 0 + 0

⇒ (1, 0).(0, 1) = 0

Multiply 2^nd row of matrix B by matching members of 2^nd column of matrix B, then sum them up.

(1, 0).(-1, 0) = (1 × -1) + (0 × 0)

⇒ (1, 0).(-1, 0) = -1 + 0

⇒ (1, 0).(-1, 0) = -1

So,

Now, compute A² – B².

Clearly,

and are not equal.

Thus, (A + B)(A – B) ≠ A² – B².

We have the matrix equation,

We need to find the value of x.

Let us compute L.H.S:

Let,

and

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

First, let us compute

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(1, x, 1).(1, 2, 15) = (1 × 1) + (x × 2) + (1 × 15)

⇒ (1, x, 1).(1, 2, 15) = 1 + 2x + 15

⇒ (1, x, 1).(1, 2, 15) = 2x + 16

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(1, x, 1).(3, 5, 3) = (1 × 3) + (x × 5) + (1 × 3)

⇒ (1, x, 1).(3, 5, 3) = 3 + 5x + 3

⇒ (1, x, 1).(3, 5, 3) = 5x + 6

Multiply 1^st row of matrix A by matching members of 3^rd column of matrix B, then sum them up.

(1, x, 1).(2, 1, 2) = (1 × 2) + (x × 1) + (1 × 2)

⇒ (1, x, 1).(2, 1, 2) = 2 + x + 2

⇒ (1, x, 1).(2, 1, 2) = x + 4

So,

Now, compute

Multiply 1^st row of matrix D by matching members of 1^st column of matrix C, then sum them up.

(2x + 16, 5x + 6, x + 4).(1, 2, x) = ((2x + 16) × 1) + ((5x + 6) × 2) + ((x + 4) × x)

⇒ (2x + 16, 5x + 6, x + 4).(1, 2, x) = (2x + 16) + (10x + 12) + (x² + 4x)

⇒ (2x + 16, 5x + 6, x + 4).(1, 2, x) = x² + 2x + 10x + 4x + 16 + 12

⇒ (2x + 16, 5x + 6, x + 4).(1, 2, x) = x² + 16x + 28

So, we have got

Now, put L.H.S = R.H.S

[x² + 16x + 28] = [0]

This means,

x² + 16x + 28 = 0

⇒ x² + 14x + 2x + 28 = 0

⇒ x(x + 14) + 2(x + 14) = 0

⇒ (x + 2)(x + 14) = 0

⇒ (x + 2) = 0 or (x + 14) = 0

⇒ x = -2 or x = -14

Thus, x = -2, -14.

We have the matrix A, such that

(i). We need to show that the matrix A satisfies the equation A² – 3A – 7I = 0.

(ii). Also, we need to find A^-1.

(i). Take L.H.S: A² – 3A – 7I

First, compute A².

A² = A.A

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix A, then sum them up.

(5, 3).(5, -1) = (5 × 5) + (3 × -1)

⇒ (5, 3).(5, -1) = 25 + (-3)

⇒ (5, 3).(5, -1) = 25 – 3

⇒ (5, 3).(5, -1) = 22

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix A, then sum them up.

(5, 3).(3, -2) = (5 × 3) + (3 × -2)

⇒ (5, 3).(3, -2) = 15 + (-6)

⇒ (5, 3).(3, -2) = 15 – 6

⇒ (5, 3).(3, -2) = 9

Multiply 2^nd row of matrix A by matching members of 1^st column of matrix A, then sum them up.

(-1, -2).(5, -1) = (-1 × 5) + (-2 × -1)

⇒ (-1, -2).(5, -1) = -5 + 2

⇒ (-1, -2).(5, -1) = -3

Multiply 2^nd row of matrix A by matching members of 2^nd column of matrix A, then sum them up.

(-1, -2).(3, -2) = (-1 × 3) + (-2 × -2)

⇒ (-1, -2).(3, -2) = -3 + 4

⇒ (-1, -2).(3, -2) = 1

Substitute values of A² and A in A² – 3A – 7I.

Also, since matrix A is of the order 2 × 2, then I will be the identity matrix of order 2 × 2 such that,

Clearly,

L.H.S = R.H.S

Thus, we have shown that matrix A satisfy A² – 3A – 7I = 0.

(ii). Now, let us find A^-1.

We know that, inverse of matrix A is A^-1 is true only when

A × A^-1 = A^-1 × A = I

Where, I = Identity matrix

We have,

A² – 3A – 7I = 0

Multiply A^-1 on both sides, we get

A^-1(A² – 3A – 7I) = A^-1 × 0

⇒ A^-1.A² – A^-1.3A – A^-1.7I = 0

⇒ A^-1.A.A – 3A^-1.A – 7A^-1.I = 0

⇒ (A^-1A)A – 3(A^-1A) – 7(A^-1I) = 0

And as A^-1A = I and A^-1I = A^-1

⇒ IA – 3I – 7A^-1 = 0

Since, IA = A

⇒ A – 3I – 7A^-1 = 0

⇒ 7A^-1 = A – 3I

[∵ ]

Thus, .

Here we have been given a matrix equation,

We need to find the matrix A.

Let matrix A be of order 2 × 2, and can be represented as

Then, we get

Take L.H.S:

So, first let us calculate

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Multiply 1^st row of matrix X by matching members of 1^st column of matrix Y, then sum them up.

(2, 1).(a, c) = (2 × a) + (1 × c)

⇒ (2, 1).(a, c) = 2a + c

Multiply 1^st row of matrix X by matching members of 2^nd column of matrix Y, then sum them up.

(2, 1).(b, d) = (2 × b) + (1 × d)

⇒ (2, 1).(b, d) = 2b + d

Multiply 2^nd row of matrix X by matching members of 1^st column of matrix Y, then sum them up.

(3, 2).(a, c) = (3 × a) + (2 × c)

⇒ (3, 2).(a, c) = 3a + 2c

Multiply 2^nd row of matrix X by matching members of 2^nd column of matrix Y, then sum them up.

(3, 2).(b, d) = (3 × b) + (2 × d)

⇒ (3, 2).(b, d) = 3b + 2d

Let X.Y = Z

Now, we need to find .

That is,

Where, let .

Multiply 1^st row of matrix Z by matching members of 1^st column of matrix Q, then sum them up.

(2a + c, 2b + d).(-3, 5) = ((2a + c) × -3) + ((2b + d) × 5)

⇒ (2a + c, 2b + d).(-3, 5) = -6a – 3c + 10b + 5d

⇒ (2a + c, 2b + d).(-3, 5) = -6a + 10b – 3c + 5d

Multiply 1^st row of matrix Z by matching members of 2^nd column of matrix Q, then sum them up.

(2a + c, 2b + d).(2, -3) = ((2a + c) × 2) + ((2b + d) × -3)

⇒ (2a + c, 2b + d).(2, -3) = 4a + 2c – 6b – 3d

⇒ (2a + c, 2b + d).(2, -3) = 4a – 6b + 2c – 3d

Multiply 2^nd row of matrix Z by matching members of 1^st column of matrix Q, then sum them up.

(3a + 2c, 3b + 2d).(-3, 5) = ((3a + 2c) × -3) + ((3b + 2d) × 5)

⇒ (3a + 2c, 3b + 2d).(-3, 5) = -9a – 6c + 15b + 10d

⇒ (3a + 2c, 3b + 2d).(-3, 5) = -9a + 15b – 6c + 10d

Multiply 2^nd row of matrix Z by matching members of 2^nd column of matrix Q, then sum them up.

(3a + 2c, 3b + 2d).(2, -3) = ((3a + 2c) × 2) + ((3b + 2d) × -3)

⇒ (3a + 2c, 3b + 2d).(2, -3) = 6a + 4c – 9b – 6d

⇒ (3a + 2c, 3b + 2d).(2, -3) = 6a – 9b + 4c – 6d

So, we have

Now, for L.H.S = R.H.S

For matrices having same order, we can write as

-6a + 10b – 3c + 5d = 1 …(i)

4a – 6b + 2c – 3d = 0 …(ii)

-9a + 15b – 6c + 10d = 0 …(iii)

6a – 9b + 4c – 6d = 1 …(iv)

We have 4 variables to find, namely, a, b, c and d; and 4 equations.

So, on adding equations (i) and (iv), we get

(-6a + 10b – 3c + 5d) + (6a – 9b + 4c – 6d) = 1 + 1

⇒ -6a + 6a + 10b – 9b – 3c + 4c + 5d – 6d = 2

⇒ 0 + b + c – d = 2

⇒ d = b + c – 2 …(v)

Now, adding equations (ii) and (iii), we get

(4a – 6b + 2c – 3d) + (-9a + 15b – 6c + 10d) = 0 + 0

⇒ 4a – 9a – 6b + 15b + 2c – 6c – 3d + 10d = 0

⇒ -5a + 9b – 4c + 7d = 0 …(vi)

On adding equations (iv) and (vi), we get

(6a – 9b + 4c – 6d) + (-5a + 9b – 4c + 7d) = 1 + 0

⇒ 6a – 5a – 9b + 9b + 4c – 4c – 6d + 7d = 1

⇒ a + 0 + 0 + d = 1

⇒ d = 1 – a …(vii)

Putting the value of d from equation (vii) in (v), we get

(1 – a) = b + c – 2

⇒ b + c – 2 – 1 = -a

⇒ b + c – 3 = -a

⇒ a = 3 – b – c …(viii)

Now, putting values of a and d from equations (vii) and (viii) in equation (iii), we get

-9(3 – b – c) + 15b – 6c + 10(1 – a) = 0

⇒ -9(3 – b – c) + 15b – 6c + 10(1 – (3 – b – c)) = 0 [∵ a = 3 – b – c]

⇒ -27 + 9b + 9c + 15b – 6c + 10(1 – 3 + b + c) = 0

⇒ -27 + 9b + 9c + 15b – 6c + 10(-2 + b + c) = 0

⇒ -27 + 9b + 9c + 15b – 6c – 20 + 10b + 10c = 0

⇒ 9b + 15b + 10b + 9c – 6c + 10c – 27 – 20 = 0

⇒ 34b + 13c – 47 = 0

⇒ 34b + 13c = 47 …(ix)

Also, putting values of a and d from equations (vii) and (viii) in equation (ii), we get

4(3 – b – c) – 6b + 2c – 3(1 – a) = 0

⇒ 12 – 4b – 4c – 6b + 2c – 3(1 – (3 – b – c)) = 0

⇒ 12 – 4b – 4c – 6b + 2c – 3(1 – 3 + b + c) = 0

⇒ 12 – 4b – 4c – 6b + 2c – 3(-2 + b + c) = 0

⇒ 12 – 4b – 4c – 6b + 2c + 6 – 3b – 3c = 0

⇒ -4b – 6b – 3b – 4c + 2c – 3c + 12 + 6 = 0

⇒ -13b – 5c + 18 = 0

⇒ 13b + 5c = 18 …(x)

On multiplying equation (ix) by 5 and equation (x) by 13, we get

(ix) ⇒ 5(34b + 13c) = 5 × 47

⇒ 170b + 65c = 235 …(xi)

(x) ⇒ 13(13b + 5c) = 13 × 18

⇒ 169b + 65c = 234 …(xii)

Subtracting equations (xi) and (xii), we get

(170b + 65c) – (169b + 65c) = 235 – 234

⇒ 170b – 169b + 65c – 65c = 1

⇒ b = 1

Putting b = 1 in equation (x), we get

13(1) + 5c = 18

⇒ 13 + 5c = 18

⇒ 5c = 18 – 13

⇒ 5c = 5

⇒ c = 1

Putting b = 1 and c = 1 in equation (viii), we get

a = 3 – b – c

⇒ a = 3 – 1 – 1

⇒ a = 3 – 2

⇒ a = 1

Putting a = 1 in equation (vii), we get

d = 1 – a

⇒ d = 1 – 1

⇒ d = 0

Thus, the matrix A is

We have,

We need to find the matrix A.

Let us see what the order of the matrices given are.

We know what order of matrix is,

If a matrix has M rows and N columns, the order of matrix is M × N.

Order of .

Number of rows = 3

⇒ M = 3

Number of column = 1

⇒ N = 1

Then, order of matrix X = M × N

⇒ Order of matrix X = 3 × 1

Order of .

Number of rows = 3

⇒ M = 3

Number of columns = 3

⇒ N = 3

Then, order of matrix Y = M × N

⇒ Order of matrix Y = 3 × 3

We must understand that, when a matrix of order 1 × 3 is multiplied to the matrix X, only then matrix Y is produced.

Let matrix A be of order 1 × 3, and can be represented as

Then, we get

Take L.H.S:

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

So, we have

Multiply 1^st row of matrix X by matching member of 1^st column of matrix A, then sum them up.

(4)(a) = 4a

Multiply 1^st row of matrix X by matching member of 2^nd column of matrix A, then sum them up.

(4)(b) = 4b

Multiply 1^st row of matrix X by matching member of 3^rd column of matrix A, then sum them up.

(4)(c) = 4c

Multiply 2^nd row of matrix X by matching member of 1^st column of matrix A, then sum them up.

(1)(a) = a

Multiply 2^nd row of matrix X by matching member of 2^nd column of matrix A, then sum them up.

(1)(b) = b

Multiply 2^nd row of matrix X by matching member of 3^rd column of matrix A, then sum them up.

(1)(c) = c

Multiply 3^rd row of matrix X by matching member of 1^st column of matrix A, then sum them up.

(3)(a) = 3a

Multiply 3^rd row of matrix X by matching member of 2^nd column of matrix A, then sum them up.

(3)(b) = 3b

Multiply 3^rd row of matrix X by matching member of 3^rd column of matrix A, then sum them up.

(3)(c) = 3c

Now, L.H.S = R.H.S

Since, the matrices on either sides are of same order, we can say that

4a = -4 …(i)

4b = 8 …(ii)

4c = 4 …(iii)

a = -1 …(iv)

b = 2 …(v)

c = 1 …(vi)

3a = -3 …(vii)

3b = 6 …(viii)

3c = 3 …(ix)

From equation (i), we can find the value of a,

4a = -4

⇒ a = -1

From equation (ii), we can find the value of b,

4b = 8

⇒ b = 2

From equation (iii), we can find the value of c,

4c = 4

⇒ c = 1

And it will satisfy other equations (iv), (v), (vi), (vii), (viii) and (ix) too.

Thus, the matrix A is

We have,

We need to verify (BA)² ≠ B²A².

Take L.H.S: (BA)²

First, compute BA.

We know what order of matrix is,

If a matrix has M rows and N columns, the order of matrix is M × N.

Order of matrix B:

Number of rows = 2

⇒ M = 2

Number of columns = 3

⇒ N = 3

Then, order of matrix = M × N

⇒ Order of matrix B = 2 × 3

Order of matrix A:

Number of rows = 3

⇒ M = 3

Number of columns = 2

⇒ N = 2

Then, order of matrix = M × N

⇒ Order of matrix A = 3 × 2

Since, in order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

So, A and B can be multiplied.

Multiply 1^st row of matrix B by matching member of 1^st column of matrix A, then sum them up.

(2, 1, 2)(3, 1, 2) = (2 × 3) + (1 × 1) + (2 × 2)

⇒ (2, 1, 2)(3, 1, 2) = 6 + 1 + 4

⇒ (2, 1, 2)(3, 1, 2) = 11

Multiply 1^st row of matrix B by matching member of 2^nd column of matrix A, then sum them up.

(2, 1, 2)(-4, 1, 0) = (2 × -4) + (1 × 1) + (2 × 0)

⇒ (2, 1, 2)(-4, 1, 0) = -8 + 1 + 0

⇒ (2, 1, 2)(-4, 1, 0) = -7

Multiply 2^nd row of matrix B by matching member of 1^st column of matrix A, then sum them up.

(1, 2, 4)(3, 1, 2) = (1 × 3) + (2 × 1) + (4 × 2)

⇒ (1, 2, 4)(3, 1, 2) = 3 + 2 + 8

⇒ (1, 2, 4)(3, 1, 2) = 13

Multiply 2^nd row of matrix B by matching member of 2^nd column of matrix A, then sum them up.

(1, 2, 4)(-4, 1, 0) = (1 × -4) + (2 × 1) + (4 × 0)

⇒ (1, 2, 4)(-4, 1, 0) = -4 + 2 + 0

⇒ (1, 2, 4)(-4, 1, 0) = -2

So,

(BA)² = (BA).(BA)

Similarly,

Take R.H.S: B²A²

Let us first compute B².

B² = B.B

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Note that in matrix B, number of columns ≠ number of rows.

This means, we can’t find B².

⇒ L.H.S ≠ R.H.S

Thus, (BA)² ≠ B²A².

We are given matrices A and B, such that

We are required to find BA and AB, if possible.

Since, in order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Let us check for BA.

If a matrix has M rows and N columns, the order of matrix is M × N.

Order of B:

Number of rows = 3

⇒ M = 3

Number of columns = 2

⇒ N = 2

Then, order of matrix B = M × N

⇒ Order of matrix B = 3 × 2

Order of A:

Number of rows = 2

⇒ M = 2

Number of columns = 3

⇒ N = 3

Then, order of matrix A = M × N

⇒ Order of matrix A = 2 × 3

Here,

Number of columns in matrix B = Number of rows in matrix A = 2

So, BA is possible.

Let us check for AB.

Here,

Number of columns in matrix A = Number of rows in matrix B = 3

So, AB is also possible.

Let us find out BA.

Multiply 1^st row of matrix B by matching members of 1^st column of matrix A, then sum them up.

(4, 1).(2, 1) = (4 × 2) + (1 × 1)

⇒ (4, 1).(2, 1) = 8 + 1

⇒ (4, 1).(2, 1) = 9

Multiply 1^st row of matrix B by matching members of 2^nd column of matrix A, then sum them up.

(4, 1).(1, 2) = (4 × 1) + (1 × 2)

⇒ (4, 1).(1, 2) = 4 + 2

⇒ (4, 1).(1, 2) = 6

Similarly, let us calculate in the matrix itself.

Now, let us find out AB.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(2, 1, 2).(4, 2, 1) = (2 × 4) + (1 × 2) + (2 × 1)

⇒ (2, 1, 2).(4, 2, 1) = 8 + 2 + 2

⇒ (2, 1, 2).(4, 2, 1) = 12

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(2, 1, 2).(1, 3, 2) = (2 × 1) + (1 × 3) + (2 × 2)

⇒ (2, 1, 2).(1, 3, 2) = 2 + 3 + 4

⇒ (2, 1, 2).(1, 3, 2) = 9

Similarly, let us calculate in the matrix itself.

Thus, and .

We know that,

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

We are given that,

A ≠ 0 and B ≠ 0

We need to show that, AB = 0.

For multiplication of A and B,

Number of columns of matrix A = Number of rows of matrix B = 2 (let)

Matrices A and B are square matrices of order 2 × 2.

For AB to become 0, one of the column of matrix A and other row of matrix B must be 0.

For example,

Check: Multiply AB.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(0, 1).(3, 0) = (0 × 3) + (1 × 0)

⇒ (0, 1).(3, 0) = 0 + 0 = 0

Similarly, let us do it for the rest of the elements.

Thus, this example justifies the criteria.

We have two matrices A and B, such that

We need to verify whether (AB)’ = B’A’.

Let us understand what a transpose is.

In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as A^T.

Take L.H.S = (AB)’

So, let us compute AB.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(2, 4, 0)(1, 2, 1) = (2 × 1) + (4 × 2) + (0 × 1)

⇒ (2, 4, 0)(1, 2, 1) = 2 + 8 + 0

⇒ (2, 4, 0)(1, 2, 1) = 10

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(2, 4, 0)(4, 8, 3) = (2 × 4) + (4 × 8) + (0 × 3)

⇒ (2, 4, 0)(4, 8, 3) = 8 + 32 + 0

⇒ (2, 4, 0)(4, 8, 3) = 40

Similarly, let us do it for the rest of the elements.

So,

Now, for transpose of AB, rows will become columns.

Now, take R.H.S = B’A’

If

Then, if (1, 4) are the elements of 1^st row, it will become elements of 1^st column, and so on.

Also,

Then, if (2, 4, 0) are the elements of 1^st row, it will become elements of 1^st column, and so on.

Now, multiply B’A’.

Multiply 1^st row of matrix B’ by matching members of 1^st column of matrix A’, then sum them up.

(1, 2, 1)(2, 4, 0) = (1 × 2) + (2 × 4) + (1 × 0)

⇒ (1, 2, 1)(2, 4, 0) = 2 + 8 + 0

⇒ (1, 2, 1)(2, 4, 0) = 10

Multiply 1^st row of matrix B’ by matching members of 2^nd column of matrix A’, then sum them up.

(1, 2, 1)(3, 9, 6) = (1 × 3) + (2 × 9) + (1 × 6)

⇒ (1, 2, 1)(3, 9, 6) = 3 + 18 + 6

⇒ (1, 2, 1)(3, 9, 6) = 27

Similarly, filling up for the rest of the elements.

⇒ L.H.S = R.H.S

Thus, (AB)’ = B’A’.

We are given with a matrix equation,

We need to find x and y.

These matrices can be added easily as they are of same order.

If two matrices are equal, then their corresponding elements are also equal.

This implies,

2x + 3y – 8 = 0 …(i)

x + 5y – 11 = 0 …(ii)

We have two variables, x and y; and two equations. It can be solved.

Rearranging equation (i), we get

2x + 3y = 8 …(iii)

Rearranging equation (ii), then multiplying it by 2 on both sides, we get

x + 5y = 11

2(x + 5y) = 2 × 11

⇒ 2x + 10y = 22 …(iv)

Subtracting equation (iii) from (iv), we get

(2x + 10y) – (2x + 3y) = 22 – 8

⇒ 2x + 10y – 2x – 3y = 14

⇒ 2x – 2x + 10y – 3y = 14

⇒ 7y = 14

⇒ y = 2

Substituting y = 2 in equation (iii), we get

2x + 3(2) = 8

⇒ 2x + 6 = 8

⇒ 2x = 8 – 6

⇒ 2x = 2

⇒ x = 1

Thus, x = 1 and y = 2.

We have the matrix equations,

…(i)

…(ii)

Subtracting equation (i) from (ii), we get

…(iii)

Adding equations (i) and (ii), we get

…(iv)

Adding equations (iii) and (iv), we get

Putting the matrix A in equation (iv), we get

Thus, and .

We have the matrices A and B, such that

We need to find matric C, such that AC = BC.

Let C be a non-zero matrix of order 2 × 1, such that

But order of C can be 2 × 1, 2 × 2, 2 × 3, 2 × 4, …

[∵ In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

∴, number of columns in matrix A = number of rows in matrix C = 2]

Take AC.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix C, then sum them up.

(3, 5)(x, y) = (3 × x) + (5 × y)

⇒ (3, 5)(x, y) = 3x + 5y

Now, take BC.

Multiply 1^st row of matrix B by matching members of 1^st column of matrix C, then sum them up.

(7, 3)(x, y) = (7 × x) + (3 × y)

⇒ (7, 3)(x, y) = 7x + 3y

And,

AC = BC

⇒ [3x + 5y] = [7x + 3y]

⇒ 3x + 5y = 7x + 3y

⇒ 7x – 3x = 5y – 3y

⇒ 4x = 2y

⇒ y = 2x

Then,

Since, C is of orders, 2 × 1, 2 × 2, 2 × 3, …

In general,

Where, k is any real number.

We need to form matrices A, B and C such that AB = AC, where A is a non-zero matrix, but B ≠ C.

Take,

First, compute AB.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(1, 0)(1, 2) = (1 × 1) + (0 × 2)

⇒ (1, 0)(1, 2) = 1 + 0

⇒ (1, 0)(1, 2) = 1

Similarly, let us do the same for other elements.

Now, let us compute AC.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix C, then sum them up.

(1, 0)(1, 2) = (1 × 1) + (0 × 2)

⇒ (1, 0)(1, 2) = 1 + 0

⇒ (1, 0)(1, 2) = 1

Similarly, let us do the same for other elements.

Clearly, AB = AC.

Thus, we have found an example that satisfy the required criteria.

We have matrices A, B and C, such that

In order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

(i). We need to verify: (AB)C = A(BC)

Take L.H.S = (AB)C

First, compute AB.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(1, 2)(2, 3) = (1 × 2) + (2 × 3)

⇒ (1, 2)(2, 3) = 2 + 6

⇒ (1, 2)(2, 3) = 8

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(1, 2)(3, -4) = (1 × 3) + (2 × -4)

⇒ (1, 2)(3, -4) = 3 – 8

⇒ (1, 2)(3, -4) = -5

Similarly, let us fill for the rest of elements.

Let .

Now, compute for DC. [∵ (AB)C = DC]

Multiply 1^st row of matrix D by matching members of 1^st column of matrix C, then sum them up.

(8, -5)(1, -1) = (8 × 1) + (-5 × -1)

⇒ (8, -5)(1, -1) = 8 + 5

⇒ (8, -5)(1, -1) = 13

Multiply 1^st row of matrix D by matching members of 2^nd column of matrix C, then sum them up.

(8, -5)(0, 0) = (8 × 0) + (-5 × 0)

⇒ (8, -5)(0, 0) = 0 + 0

⇒ (8, -5)(0, 0) = 0

Similarly, let us fill for the rest of elements.

So,

Take R.H.S: A(BC)

First, compute BC.

Multiply 1^st row of matrix B by matching members of 1^st column of matrix C, then sum them up.

(2, 3)(1, -1) = (2 × 1) + (3 × -1)

⇒ (2, 3)(1, -1) = 2 – 3

⇒ (2, 3)(1, -1) = -1

Multiply 1^st row of matrix B by matching members of 2^nd column of matrix C, then sum them up.

(2, 3)(0, 0) = (2 × 0) + (3 × 0)

⇒ (2, 3)(0, 0) = 0 + 0

⇒ (2, 3)(0, 0) = 0

Similarly, let us fill for the rest of the elements.

Let .

Now, compute for AE.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix E, then sum them up.

(1, 2)(-1, 7) = (1 × -1) + (2 × 7)

⇒ (1, 2)(-1, 7) = -1 + 14

⇒ (1, 2)(-1, 7) = 13

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix E, then sum them up.

(1, 2)(0, 0) = (1 × 0) + (2 × 0)

⇒ (1, 2)(0, 0) = 0 + 0

⇒ (1, 2)(0, 0) = 0

Similarly, repeat the step for the other elements.

So,

Thus, (AB)C = A(BC).

(ii). We need to verify: A(B + C) = AB + AC

Take L.H.S: A(B + C)

Add B + C.

Let B + C = F, such that

Now, multiply A and F.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix F, then sum them up.

(1, 2)(3, 2) = (1 × 3) + (2 × 2)

⇒ (1, 2)(3, 2) = 3 + 4

⇒ (1, 2)(3, 2) = 7

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix F, then sum them up.

(1, 2)(3, -4) = (1 × 3) + (2 × -4)

⇒ (1, 2)(3, -4) = 3 – 8

⇒ (1, 2)(3, -4) = -5

Similarly, repeat the steps for the other elements.

So,

Now, take R.H.S: AB + AC

Compute AB.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(1, 2)(2, 3) = (1 × 2) + (2 × 3)

⇒ (1, 2)(2, 3) = 2 + 6

⇒ (1, 2)(2, 3) = 8

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix B, then sum them up.

(1, 2)(3, -4) = (1 × 3) + (2 × -4)

⇒ (1, 2)(3, -4) = 3 – 8

⇒ (1, 2)(3, -4) = -5

Similarly, repeat the steps for the other elements.

So,

Now, compute AC.

Multiply 1^st row of matrix A by matching members of 1^st column of matrix C, then sum them up.

(1, 2)(1, -1) = (1 × 1) + (2 × -1)

⇒ (1, 2)(1, -1) = 1 – 2

⇒ (1, 2)(1, -1) = -1

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix C, then sum them up.

(1, 2)(0, 0) = (1 × 0) + (2 × 0)

⇒ (1, 2)(0, 0) = 0 + 0

⇒ (1, 2)(0, 0) = 0

Similarly, repeat the steps for the other elements.

So,

Adding AB + AC.

Matrices of same order can be added or subtracted.

So, clearly L.H.S = R.H.S.

Thus, A(B + C) = AB + AC.

Given: We have matrices P and Q, such that

To Prove:

Proof: First, we shall compute PQ.

Since, in order to multiply two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Order of P = 3 × 3

And order of Q = 3 × 3

Number of columns of matrix P = Number of rows of matrix Q = 3

So, P and Q can be multiplied.

So, multiply 1^st row of matrix P by matching members of 1^st column of matrix Q, then sum them up.

(x, 0, 0)(a, 0, 0) = (x × a) + (0 × 0) + (0 × 0)

⇒ (x, 0, 0)(a, 0, 0) = xa

Multiply 1^st row of matrix P by matching members of 2^nd column of matrix Q, then sum them up.

(x, 0, 0)(0, b, 0) = (x × 0) + (0 × b) + (0 × 0)

⇒ (x, 0, 0)(0, b, 0) = 0

Similarly, repeat the steps to find other elements.

So,

…(i)

Now, we shall compute QP.

Multiply 1^st row of matrix Q by matching members of 1^st column of matrix P, then sum them up.

(a, 0, 0)(x, 0, 0) = (a × x) + (0 × 0) + (0 × 0)

⇒ (a, 0, 0)(x, 0, 0) = xa + 0 + 0

⇒ (a, 0, 0)(x, 0, 0) = xa

Similarly, repeat the steps to find other elements.

So,

Thus, .

We are given with a matrix equation,

We need to find A.

Take L.H.S:

Let us solve , where

Then,

Order of X = 1 × 3

Order of Y = 3 × 3

Then, resulting order of matrix Z(say) = 1 × 3 [Let Z = XY]

Multiply 1^st row of matrix X by matching members of 1^st column of matrix Y, then sum them up.

(2, 1, 3)(-1, -1, 0) = (2 × -1) + (1 × -1) + (3 × 0)

⇒ (2, 1, 3)(-1, -1, 0) = -2 – 1 + 0

⇒ (2, 1, 3)(-1, -1, 0) = -3

Multiply 1^st row of matrix X by matching members of 2^nd column of matrix Y, then sum them up.

(2, 1, 3)(0, 1, 1) = (2 × 0) + (1 × 1) + (3 × 1)

⇒ (2, 1, 3)(0, 1, 1) = 0 + 1 + 3

⇒ (2, 1, 3)(0, 1, 1) = 4

Multiply 1^st row of matrix X by matching members of 3^rd column of matric Y, then sum them up.

(2, 1, 3)(-1, 0, 1) = (2 × -1) + (1 × 0) + (3 × 1)

⇒ (2, 1, 3)(-1, 0, 1) = -2 + 0 + 3

⇒ (2, 1, 3)(-1, 0, 1) = 1

So,

Now, multiplying Z by .

Order of Z = 1 × 3

Order of Q = 3 × 1

Then, order of the resulting matrix = 1 × 1

Multiply 1^st row of matrix Z by matching members of 1^st column of matrix Q, then sum them up.

(-3, 4, 1)(1, 0, -1) = (-3 × 1) + (4 × 0) + (1 × -1)

⇒ (-3, 4, 1)(1, 0, -1) = -3 + 0 – 1

⇒ (-3, 4, 1)(1, 0, -1) = -4

Now, since

Thus,

A = [-4]

We are given the matrices A, B and C, such that

We need to verify that, A(B + C) = AB + AC.

Take L.H.S: A(B + C)

Solving (B + C).

These matrices can be added as they have same order.

Now, multiply A by (B + C).

Let (B + C) = D.

We have,

AD = A(B + C)

Order of A = 1 × 2

Order of D = 2 × 3

Then, order of resulting matrix = 1 × 3

Multiply 1^st row of matrix A by matching members of 1^st column of matrix D, then sum them up.

(2, 1)(4, 9) = (2 × 4) + (1 × 9)

⇒ (2, 1)(4, 9) = 8 + 9

⇒ (2, 1)(4, 9) = 17

Multiply 1^st row of matrix A by matching members of 2^nd column of matrix D, then sum them up.

(2, 1)(5, 7) = (2 × 5) + (1 × 7)

⇒ (2, 1)(5, 7) = 10 + 7

⇒ (2, 1)(5, 7) = 17

Multiply 1^st row of matrix A by matching members of 3^rd column of matrix D, then sum them up.

(2, 1)(5, 8) = (2 × 5) + (1 × 8)

⇒ (2, 1)(5, 8) = 10 + 8

⇒ (2, 1)(5, 8) = 18

So,

Now, take R.H.S: AB + AC

Let us compute AB.

Order of A = 1 × 2

Order of B = 2 × 3

Then, order of AB = 1 × 3

Multiply 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(2, 1)(5, 8) = (2 × 5) + (1 × 8)

⇒ (2, 1)(5, 8) = 10 + 8

⇒ (2, 1)(5, 8) = 18

Similarly, repeat steps to find the rest of the elements.

Now, let us compute AC.

Order of AC = 1 × 3

Multiply 1^st row of matrix A by matching members of 1^st column of matrix C, then sum them up.

(2, 1)(-1, 1) = (2 × -1) + (1 × 1)

⇒ (2, 1)(-1, 1) = -2 + 1

⇒ (2, 1)(-1, 1) = -1

Similarly, repeat steps to find the rest of the elements.

Add, AB + AC.

Thus,

A(B + C) = AB + AC.

We are given with matrix A, such that

We need to verify A² + A = A(A + I).

Take L.H.S: A² + A.

Solve for A².

A² = A.A

Multiply 1^st row of matrix A by matching members of 1^st column of matrix A, then sum them up.

(1, 0, -1)(1, 2, 0) = (1 × 1) + (0 × 2) + (-1 × 0)

⇒ (1, 0, -1)(1, 2, 0) = 1 + 0 + 0

⇒ (1, 0, -1)(1, 2, 0) = 1

Similarly, repeat steps to fill for the other elements.

Now, add A² and A,

Take R.H.S: A(A + I)

First, let us solve for (A + I).

Multiply (A + I) from A.

Since, L.H.S = R.H.S.

Thus, (A² + A) = A(A + I).

We are given with matrices A and B, such that

(i). We need to verify that, (A’)’ = A.

Take L.H.S: (A’)’

In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as A^T or A’.

So, in transpose of a matrix,

Rows of matrix becomes columns of the same matrix.

So,

If ,

(0, -1, 2) and (4, 3, -4) are 1^st and 2^nd rows respectively, will become 1^st and 2^nd columns respectively.

Then

Also, if ,

Similarly, (0, 4), (-1, 3) and (2, -4) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Then

Note, that

Thus, verified that (A’)’ = A.

(ii). We need to verify that, (AB)’ = B’A’.

Take L.H.S: (AB)’

Compute AB.

Order of A = 2 × 3

Order of B = 3 × 2

Then, order of AB = 2 × 2

Multiplying 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(0, -1, 2)(4, 1, 2) = (0 × 4) + (-1 × 1) + (2 × 2)

⇒ (0, -1, 2)(4, 1, 2) = 0 – 1 + 4

⇒ (0, -1, 2)(4, 1, 2) = 3

Similarly, repeat the process to find the other elements.

Transpose of AB is (AB)’.

(3, 9) and (11, -15) are 1^st and 2^nd rows respectively, will become 1^st and 2^nd columns respectively.

Take R.H.S: B’A’

If ,

(4, 0), (1, 3) and (2, 6) are 1^st, 2^nd and 3^rd rows of matrix B respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Also, if ,

(0, -1, 2) and (4, 3, -4) are 1^st and 2^nd rows respectively, will become 1^st and 2^nd columns respectively.

Multiply B’ by A’.

Order of B’ = 2 × 3

Order of A’ = 3 × 2

Then, order of B’A’ = 2 × 2

Multiply 1^st row of matrix B’ by matching members of 1^st column of matrix A’, then sum them up.

(4, 1, 2)(0, -1, 2) = (4 × 0) + (1 × -1) + (2 × 2)

⇒ (4, 1, 2)(0, -1, 2) = 0 – 1 + 4

⇒ (4, 1, 2)(0, -1, 2) = 3

Similarly, repeat the same steps to find out other elements.

Since, L.H.S = R.H.S.

Thus, (AB)’ = B’A’.

(iii). We need to verify that, (kA)’ = kA’.

Take L.H.S: (kA)’

We know that,

Multiply k on both sides, (k is a scalar quantity)

Now, to find transpose of kA,

(0, -k, 2k) and (4k, 3k, -4k) are 1^st and 2^nd rows of matrix kA respectively, will become 1^st and 2^nd columns respectively.

Take R.H.S: kA’

If

Then, for transpose of A,

(0, -1, 2) and (4, 3, -4) are 1^st and 2^nd rows of matrix A respectively, will become 1^st and 2^nd columns respectively.

Multiply k on both sides,

Note that, L.H.S = R.H.S.

Thus, (kA)’ = kA’.

We are given matrices A and B, such that

In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as A^T or A’.

So, in transpose of a matrix,

Rows of matrix becomes columns of the same matrix.

(i). We need to verify that, (2A + B)’ = 2A’ + B’.

Take L.H.S: (2A + B)’

Substitute the matrices A and B, in (2A + B)’.

For transpose of (2A + B),

(3, 6), (14, 6) and (17, 15) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Take R.H.S: 2A’ + B’

If ,

(1, 2), (4, 1) and (5, 6) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Multiply both sides by 2,

Also,

If .

(1, 2), (6, 4) and (7, 3) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Now, add 2A’ and B’.

Since, L.H.S = R.H.S

Thus, (2A + B)’ = 2A’ + B’.

(ii). We need to verify that, (A – B)’ = A’ – B’.

Take L.H.S: (A – B)’

Substitute the matrices A and B in (A – B)’.

To find transpose of (A – B),

(0, 0), (-2, -3) and (-2, 3) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Take R.H.S: A’ – B’

If ,

(1, 2), (4, 1) and (5, 6) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Also,

If ,

(1, 2), (6, 4) and (7, 3) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Subtract B’ from A’,

Since, L.H.S = R.H.S

Thus, (A – B)’ = A’ – B’.

We must understand,

In linear algebra, a symmetric matrix is a square matrix that is equal to its transpose. Formally, because equal matrices have equal dimensions, only square matrices can be symmetric.

And we know that, transpose of AB is given by

(AB)’ = B’A’

Using this result, take transpose of A’A.

Transpose of A’A = (A’A)^T = (A’A)’

Using, transpose of A’A = (A’A)’

⇒ (A’A)’ = A’(A’)’

And also,

(A’)’ = A

So,

(A’A)’ = A’A

Since, (A’A)’ = A’A

This means, A’A is symmetric matrix for any matrix A.

Now, take transpose of AA’.

Transpose of AA’ = (AA’)’

⇒ (AA’)’ = (A’)’A’ [∵ (AB)’ = B’A’]

⇒ (AA’)’ = AA’ [∵ (A’)’ = A]

Since, (AA’)’ = AA’

This means, AA’ is symmetric matrix for any matrix A.

Thus, A’A and AA’ are symmetric matrix for any matrix A.

We are given that,

A and B are square matrices of the order 3 × 3.

We need to check whether (AB)² = A²B² is true or not.

Take (AB)².

(AB)² = (AB)(AB)

[∵ A and B are of order (3 × 3) each, A and B can be multiplied; A and B be any matrices of order (3 × 3)]

⇒ (AB)² = ABAB

[∵ (AB)(AB) = ABAB]

⇒ (AB)² = AABB

[∵ ABAB = AABB; as A can be multiplied with itself and B can be multiplied by itself]

⇒ (AB)² = A²B²

So, note that, (AB)² = A²B² is possible.

But this is possible if and only if BA = AB.

And BA = AB is always true whenever A and B are square matrices of any order. And for BA = AB,

(AB)² = A²B²

By matrix multiplication we can write:

(A + B)² = (A+B)(A+B) = A² + AB + BA + B²

We know that matrix multiplication is not commutative but it is given that : AB = BA

∴ (A + B)² = A² + AB + AB + B²

⇒ (A + B)² = A² + 2AB + B² …proved

Given A = B = and C =

LHS = A + (B + C) =

⇒ LHS =

⇒ LHS =

RHS = (A + B) + C =

⇒ RHS =

⇒ RHS =

Clearly LHS = RHS =

Hence,

A + (B + C) = (A + B) + C …proved

To prove: A(BC) = (AB)C

LHS = A(BC) =

⇒ LHS =

⇒ LHS =

⇒ LHS =

RHS = (AB)C =

Performing matrix multiplication as done for LHS

⇒ RHS =

⇒ RHS =

Clearly LHS = RHS =

∴ A(BC) = (AB)C …proved

To prove: (a + b)B = aB + bB

Given, a = 4 and b = -2

LHS = (4+(-2))B =

RHS = aB + bB =

⇒ RHS =

Clearly LHS = RHS =

Hence,

(a + b)B = aB + bB …proved

To prove: a(C – A) = aC –aA

As, LHS = a(C – A) = 4

⇒ LHS = 4

RHS = aC – aA =

⇒ aC – aA =

Clearly LHS = RHS =

Hence,

a(C – A) = aC –aA …proved

To prove: (A^T)^T = A

As transpose of a matrix is obtained by interchanging rows with respective columns.

LHS = (A^T)^T = = RHS

Hence, proved.

a) To prove: (bA)^T = bA^T

As, LHS = (bA)^T = (-2A)^T

⇒ LHS =

Similarly,

RHS =

Clearly LHS = RHS =

Hence, (bA)^T = bA^T …proved

b) To prove: (AB)^T = B^TA^T

Clearly, LHS = (AB)^T =

First multiplying the matrix and then taking the transpose.

∴ LHS =

⇒ LHS =

∴ LHS =

As RHS = B^TA^T

We will first take transpose of matrices and then multiply

RHS =

⇒ RHS =

Clearly, LHS = RHS =

Hence (AB)^T = B^TA^T …proved

c) To prove: (A – B)C = AC – BC

As, LHS = (A – B)C

Putting the values of A,B and C and multiplying by rule of matrix multiplication.

LHS =

⇒ LHS =

RHS = AC – BC =

⇒ RHS =

Clearly, LHS = RHS =

Hence (A-B)C = AC - BC…proved

To Prove: (A – B)^T = A^T - B^T

LHS = (A – B)^T =

RHS = A^T – B^T =

∴ RHS =

Clearly, LHS = RHS =

Hence (A-B)^T = A^T – B^T …proved

As A =

∴ A² =

By matrix multiplication:

A² =

⇒ A² =

As we know that:

2 sin θ cos θ = sin 2θ and cos² θ – sin² θ = cos 2θ

∴ A² = …Hence proved

As, LHS = (A + B)² =

⇒ LHS =

By matrix multiplication we can write LHS as –

LHS =

⇒ LHS =

Given x² = -1

∴ LHS =

RHS = A² + B² =

⇒ RHS =

By matrix multiplication we can write-

RHS =

Given x² = -1

∴ RHS =

Clearly RHS = LHS =

Hence, (A + B)² = A² + B² …proved

We need to prove that: A² = I =

∵ A =

∴ A² =

By matrix multiplication we have-

A² =

⇒ A² =

∴ A² =

Hence Verified

By principle of mathematical induction we say that if a statement P(n) is true for n = 1 and if we assume P(k) to be true for some random natural number k and usnig it if we prove P(k+1) to be true we can say that P(n) is true for all natural numbers.

We are given to prove that (A’)ⁿ = (Aⁿ)’.

Let P(n) be the statement : (A’)ⁿ = (Aⁿ)’.

Clearly, P(1): (A’)¹ = (A¹)’

⇒ P(1) : A’ = A’

⇒ P(1) is true

Let P(k) be true.

∴ (A’)^k = (A^k)’ …(1)

Let’s take P(k+1) now:

∵ (A^k+1)’ = (A^kA)’

We know that by properties of transpose of a matrix:

(AB)^T = B^TA^T

∴ (A^kA)’ = A’(A^k)’ = A’(A’)^k = (A’)^k+1

Thus, (A^k+1)’ = (A’)^k+1

∴ P(k+1) is true.

Hence,

We can say that: (A’)ⁿ = (Aⁿ)’ is true for all n ∈ N.

Let A =

To apply elementary row transformations we write:

A = IA where I is the identity matrix

We proceed with operations in such a way that LHS becomes I and the transformations in I give us a new matrix such that

I = XA

And this X is called inverse of A = A^-1

So we have:

Applying R₂→ R₂ + 5R₁

⇒

Applying R₂→ (1/22)R₂

⇒

Applying R₁→ R₁ – 3R₂

⇒

As we got Identity matrix in LHS.

∴ A^-1 =

Let B =

To apply elementary row transformations we write:

B = IB where I is the identity matrix

We proceed with operations in such a way that LHS becomes I and the transformations in I give us a new matrix such that

I = XB

And this X is called inverse of B = B^-1

So we have:

Applying R₂→ R₂ + 2R₁

⇒

As we got all zeroes in one of the row of matrix in LHS.

So by any means we can make identity matrix in LHS.

∴ inverse of B does not exist.

B^-1 does not exist. …ans

Given,

As the 2 matrices are equal. So corresponding elements of both the matrix must also hold the equality.

∴ xy = 8 ; w = 4 ; z + 6 = 0 and x + y = 6

Hence, we get:

w = 4

z = -6

∵ x + y = 6

⇒ y = 6 – x

∴ x(6-x) = 8

⇒ x² – 6x + 8 = 0

⇒ x² – 4x – 2x + 8 = 0

⇒ x(x – 4) – 2(x – 4) = 0

⇒ (x – 2)(x – 4) = 0

⇒ x = 2 or x = 4

When x = 2 ; y = 4

And when x = 4 ; y = 2

Thus,

x = 2 or 4 ; y = 4 or 2 ; z = -6 and w = 4 …ans

Given that:

3A + 5B + 2C = O = null matrix

C = ?

As,

⇒

⇒

∴ 2C =

⇒ 2C =

∴ C = …ans

Given, A =

∴ A² =

By matrix multiplication we can write:

A² =

⇒ A² = …(1)

As we have to find: A² – 5A – 14I

∴ A² – 5A – 14I =

⇒ A² – 5A – 14I =

⇒ A² – 5A – 14I =

⇒ A² – 5A – 14I = = O

We need to find value of A³ using the above equation:

Now we have,

A² – 5A – 14I = O

⇒ A² = 5A + 14I

Multiplying with A both sides

⇒ A².A = 5A.A + 14IA

⇒ A³ = 5A² + 14A

Using equation 1 we get:

⇒ A³ =

⇒ A³ =

⇒ A³ =

Given,

We need to find the value of a, b, c and d.

As,

⇒

As both matrices are equal so their corresponding elements must also be equal.

∴ 3a = a + 4

⇒ 2a = 4

⇒ a = 2

Similarly,

3b = 6 + a + b

⇒ 2b = 6 + a

As from above a = 2

∴ 2b = 6+2 = 8

⇒ b = 4

Also 3d = 2d + 3

⇒ d = 3

And,

3c = -1 + c + d

⇒ 2c = d – 1

⇒ 2c = 3-1

⇒ c = 2/2 = 1

Thus a = 2, b = 4, c = 1 and d = 3.

Given,

As A is multiplied with a matrix of order 3×2 and gives a resultant matrix of order 3×3

For matrix multiplication to be possible A must have 2 rows and as resultant matrix is of 3^rd order A must have 3 columns

∴ A is matrix of order 2×3

Let A = where a, b, c, d, e and f are unknown variables.

∴

∴ By matrix multiplication we have-

By equating the elements of 2 equal matrices we get-

a = 1 ; b = -2 and c = -5

also,

2a – d = -1 ⇒ d = 2a + 1 = 2 + 1 = 3

∴ d = 3

2b – e = -8 ⇒ e = 2b + 8 = -4 + 8 = 4

∴ e = 4

Similarly, f = 2c + 10 = 0

∴ A =

Given,

∵ A² = A.A

⇒ A² =

By matrix multiplication, we get

A² =

⇒ A² =

∴ A² + 2A + 7I =

⇒ A² + 2A + 7I =

⇒ A² + 2A + 7I =

⇒ A² + 2A + 7I = …ans

Given, A =

We know that transpose of a matrix is obtained by interchanging rows with respective columns.

∴ A’ =

Inverse of a matrix A = A^-1 =

Clearly |A| =

∴ |A| = {using trigonometric identity}

Adj(A) is given by the transpose of the cofactor matrix.

∴ adj(A) =

∴ A^-1 =

According to question:

A’ = A^-1

∴

As both the matrices are equal irrespective of the value of α.

∴ α can be any real number …ans

A matrix is said to be skew-symmetric if A = -A’

Let, A =

As, A is skew symmetric matrix.

∴ A = -A’

⇒

⇒

⇒

Equating the respective elements of both matrices we get-

a = -2 ; c = -3 ; b = -b ⇒ 2b = 0 ⇒ b = 0

Thus, we have-

a = -2 , b = 0 and c = -3 …ans

Given,

P(x) = …(1)

P(y) =

∴ P(x).P(y) =

⇒ P(x).P(y) =

We know that-

cos x cos y + sin x sin y = cos (x – y)

cos x sin y + sin x cos y = sin (x + y)

and cos x cos y – sin x sin y = cos (x + y)

⇒ P(x).P(y) =

In comparison with equation 1 we can say that:

…(2)

∴ P(x).P(y) = P(x + y)

Similarly, we can show for P(y).P(x):

P(y).P(x) =

By matrix multiplication, we have –

P(y).P(x) =

⇒ P(y).P(x) =

⇒ P(y).P(x) = …(3)

∴ From equation 2 and 3:

P(x).P(y) = P(y).P(x) = P(x + y) …ans

Given that,

A² = A

∵ (a+b)³ = a³ + b³ + 3a²b + 3ab²

As, (I + A)³ = I³ + A³ + 3I²A + 3IA²

∵ I is an identity matrix.

∴ I³ = I² = I

∴ (I + A)³ = I + A³ + 3IA + 3IA

As, I is an identity matrix.

∴ IA = AI = A

⇒ (I + A)³ = I + A³ + 6IA

∵ A² = A

⇒ (I + A)³ = I + A².A + 6A

⇒ (I + A)³ = I + A.A + 6A

⇒ (I + A)³ = I + A² + 6A

⇒ (I + A)³ = I + A + 6A = I + 7A

Hence,

(I + A)³ = I + 7A …proved

A matrix is said to be skew-symmetric if A = -A’

Given, B is a skew-symmetric matrix.

∴ B = -B’

Let C = A’ BA …(1)

We have to prove C is skew-symmetric.

To prove: C = -C’

As C’ = (A’BA)’

We know that: (AB)’ = B’A’

⇒ C’ = (A’BA)’ = A’B’(A’)’

⇒ C’ = A’B’A {∵ (A’)’ = A}

⇒ C’ = A’(-B)A

⇒ C’ = -A’BA …(2)

From equation 1 and 2:

We have,

C’ = -C

Thus we say that C = A’ BA is a skew-symmetric matrix.

By principle of mathematical induction we say that if a statement P(n) is true for n = 1 and if we assume P(k) to be true for some random natural number k and usnig it if we prove P(k+1) to be true we can say that P(n) is true for all natural numbers.

We are given to prove that (AB)ⁿ = AⁿBⁿ

Let P(n) be the statement : (AB)ⁿ = AⁿBⁿ

Clearly, P(1): (AB)¹ = A¹B¹

⇒ P(1) : AB = AB

⇒ P(1) is true

Let P(k) be true.

∴ (AB)^k = A^kB^k …(1)

Let’s take P(k+1) now:

∵ (AB)^k+1 = (AB)^k(AB)

⇒ (AB)^k+1 = A^kB^k(AB)

NOTE: As we know that Matrix multiplication is not commutative. So we can’t write directly that A^kB^k(AB) = A^k+1B^k+1

But we are given that AB = BA

∴ (AB)^k+1 = A^kB^k(AB)

⇒ (AB)^k+1 = A^kB^k-1(BAB)

As AB = BA

⇒ (AB)^k+1 = A^kB^k-1(ABB)

⇒ (AB)^k+1 = A^kB^k-1(AB²)

⇒ (AB)^k+1 = A^kB^k-2(BAB²)

⇒ (AB)^k+1 = A^kB^k-2(ABB²)

⇒ (AB)^k+1 = A^kB^k-2(AB³)

We observe that one power of B is decreasing while other is increasing. After certain repetitions decreasing power of B will become I

And at last step:

⇒ (AB)^k+1 = A^kI(AB^k+1)

⇒ (AB)^k+1 = A^kAB^k+1

⇒ (AB)^k+1 = A^k+1B^k+1

Thus P(k+1) is true when P(k) is true.

∴ (AB)ⁿ = Aⁿ Bⁿ ∀ n ∈ N when AB = BA.

Given,

We need to find x, y and z such that A’ = A^-1

If A’ = A^-1

Pre-multiplying A on both sides:

AA’ = AA^-1

⇒ AA’ = I where I is the identity matrix.

∴

⇒

By matrix multiplication we have:

⇒

On equating the corresponding elements of matrix.

We need basically 3 equations as we have 3 variables to solve for. You can pick any three elements and equate them.

We have:

4y² + z² = 1 …(1)

x² + y² + z² = 1 …(2)

2y² – z² = 0 …(3)

Adding equation 2 and 3:

6y² = 1

⇒ y² = 1/6

∴

From equation 3:

Z² = 2y²

⇒ z² = 2(1/6)

∴ z² = 1/3

∴

From equation 2:

x² = 1 – y² – z²

⇒ x² = 1 – (1/6) – (1/3)

⇒ x² = 1 – 1/2 = 1/2

∴

Thus,

; and

Let A =

To apply elementary row transformations we write:

A = IA where I is the identity matrix

We proceed with operations in such a way that LHS becomes I and the transformations in I give us a new matrix such that

I = XA

And this X is called inverse of A = A^-1

Note: Never apply row and column transformations simultaneously over a matrix.

So we have:

Applying R₂→ R₂ + R₁

⇒ =

Applying R₃→ R₃ - R₂

⇒ =

Applying R₁→ R₁ + R₂

⇒ =

Applying R₂→ R₂ - 3R₁

=

Applying R₃→ (-1)R₃

⇒ =

Applying R₁→ R₁ + 10R₃ and R₂→ R₂ + 17R₃

⇒ =

Applying R₁→ (-1)R₁ and R₂→ (-1)R₂

⇒ =

As we got Identity matrix in LHS.

∴ A^-1 =

Let A =

To apply elementary row transformations we write:

A = IA where I is the identity matrix

We proceed with operations in such a way that LHS becomes I and the transformations in I give us a new matrix such that

I = XA

And this X is called inverse of A = A^-1

Note: Never apply row and column transformations simultaneously over a matrix.

So we have:

Applying R₂→ R₂ + R₃

⇒

Applying R₁→ R₁ - 2R₃

⇒

Applying R₂→ R₁ + R₂

⇒

As second row of LHS contains all zeros, So by anyhow we are never going to get Identity matrix in LHS.

∴ Inverse of A does not exist.

A^-1 does not exist. …ans

Let A =

To apply elementary row transformations we write:

A = IA where I is the identity matrix

We proceed with operations in such a way that LHS becomes I and the transformations in I give us a new matrix such that

I = XA

And this X is called inverse of A = A^-1

Note: Never apply row and column transformations simultaneously over a matrix.

So we have:

Applying R₂→ R₂ – (5/2)R₁

⇒

Applying R₃→ R₃ - R₂

⇒

Applying R₁→ R₁ + R₂

⇒ =

Applying R₂→ R₂ - 5R₃

=

Applying R₁→ R₁ + 2R₃

⇒ =

Applying R₁→ (1/2)R₁ and R₃→ 2R₃

⇒ =

As we got Identity matrix in LHS.

∴ A^-1 =

As P has equal number of rows and columns and thus it matches with the definition of square matrix.

The given matrix does not satisfy the definition of unit and diagonal matrices.

∴ Option (A) is the only correct answer.

As matrix has total 3× 3 = 9 elements.

As each element can take 2 values (0 or 2)

∴ By simple counting principle we can say that total number of possible matrices = total number of ways in which 9 elements can take possible values = 2⁹ = 512

Clearly it matches with option D.

∴ option (D) is the only correct answer.

Given,

By equality of two matrices, we have-

4x = x + 6

⇒ 3x = 6 ⇒ x = 2

Also, 2x + y = 7

⇒ y = 7 – 2x = 7 – 4 = 3

∴ y = 3

As only option (B) matches with our answer.

∴ Option(B) is the correct answer.

A very basic idea of Inverse trigonometric function is required to solve the problem.

cos^-1 x + sin^-1 x = π/2 and cot^-1 x + tan^-1 x = π/2

As,

And

∴ A – B =

⇒ A – B =

∴ A – B =

Clearly It matches with option (D)

∴ option(D) is the only correct answer.

As order of A is 3 × m and order of B is 3 × n

As m = n. So, order of A and B is same = 3 × m

∴ subtraction is possible.

And (5A – 3B) also has same order.

Let A =

∴ A² =

By matrix multiplication:

⇒ A² = which matches with option (D)

∴ Option (D) is the correct answer.

According to question:

a₁₁ = 0 , a₁₂ = 1 , a₂₁ = 1 and a₂₂ = 0

∴ A =

∴ A² =

By matrix multiplication:

⇒ A² = which matches with option (A)

∴ Option (A) is the correct answer.

Let A =

Clearly,

A’ =

As, A^T = A

∴ It is symmetric matrix.

∴ Option(B) is the correct answer.

Let A =

Clearly,

A’ =

As, A^T = -A

∴ It is skew - symmetric matrix.

∴ Option(C) is the correct answer.

As AB’ is defined. So, B’ must have n rows.

∴ B has n columns.

And, B’A is also defined. As, A’ has order n × m

∴ B’A to exist B must have m rows.

∴ m × n is the order of B.

Option (D) is the correct answer.

Let C = (AB’ – BA’)

C’ = (AB’ – BA’)’

⇒ C’ = (AB’)’ – (BA’)’

⇒ C’ = (B’)’A’ – (A’)’B’

⇒ C’ = BA’ – AB’

⇒ C’ = -C

∴ C is a skew-symmetric matrix.

Clearly Option (A) matches with our deduction.

∴ Option (A) is the correct.

As, (A – I)³ + (A + I)³

Use a³ + b³ = (a + b)(a² + ab + b²)

Also A² = I

∴ then (A – I)³ + (A + I)³ – 7A = 2A + 6A – 7A = A

Clearly our answer matches with option (A)

∴ option (A) is the correct answer.

For any two matrix:

Not always option A , B and C are true.

∴ Option (D) is the only suitable answer

For column transformation, we operate the post matrix.

As,

Applying C₂→ C₂ — 2C₁

∴

Clearly, it matches with option (D).

∴ Option (D) is the correct answer.

Elementary row transformation is applied on the first matrix of RHS.

Applying R₁→ R₁ — 3R₂ we have -

⇒

Clearly it matches with option (A)

∴ Option (A) is the correct answer.

A Zero matrix

∴ Let A be the symmetric and skew symmetric matrix.

⇒ A’=A (Symmetric)

⇒ A’=-A (Skew-Symmetric)

Considering the above two equations,

⇒ A=-A

⇒ 2A=0

⇒ A=0 (A Zero Matrix)

Hence Zero matrix is both symmetric and skew symmetric matrix.

A Zero matrix

∴ Let A be the symmetric and skew symmetric matrix.

⇒ A’=A (Symmetric)

⇒ A’=-A (Skew-Symmetric)

Considering the above two equations,

⇒ A=-A

⇒ 2A=0

⇒ A=0 (A Zero Matrix)

Hence Zero matrix is both symmetric and skew symmetric matrix.

A skew symmetric matrix

∴ Let A and B are two skew symmetric matrices.

⇒ A’=-A ..(1)

⇒ B’=-B ..(2)

Now Let A+B=C ..(3)

⇒ C’=(A+B)’=A’+B’

⇒ A’+B’=(-A)+(-B)

⇒ (-A)+(-B)=-(A+B)=-C

⇒ C’=-C (Skew Symmetric matrix)

A skew symmetric matrix

∴ Let A and B are two skew symmetric matrices.

⇒ A’=-A ..(1)

⇒ B’=-B ..(2)

Now Let A+B=C ..(3)

⇒ C’=(A+B)’=A’+B’

⇒ A’+B’=(-A)+(-B)

⇒ (-A)+(-B)=-(A+B)=-C

⇒ C’=-C (Skew Symmetric matrix)

-1

The negative of a matrix is obtained by multiplying it by -1.

For example:

Let A =

So

= - A

-1

The negative of a matrix is obtained by multiplying it by -1.

For example:

Let A =

So

= - A

The null matrix is the one in which all elements are zero.

If we want to make A = a null matrix we need to multiply it by 0.

0A =

The product of any matrix by the scalar 0 is the null matrix.

Rectangular Matrix

As a square matrix is the one in which there are same number of rows and columns.

Eg: A =

Here there are 2 rows and 2 columns.

The matrix which is not square is called rectangular matrix as it does not have same number of rows and columns.

Eg

Here number of rows are 2 and columns are 3.

Rectangular Matrix

As a square matrix is the one in which there are same number of rows and columns.

Eg: A =

Here there are 2 rows and 2 columns.

The matrix which is not square is called rectangular matrix as it does not have same number of rows and columns.

Eg

Here number of rows are 2 and columns are 3.

Distributive

⇒ Matrix multiplication is distributive over addition.

i.e A(B+C)=AB+AC

and (A+B)C=AC+BC

Distributive

⇒ Matrix multiplication is distributive over addition.

i.e A(B+C)=AB+AC

and (A+B)C=AC+BC

A³ is Also a symmetric matrix.

Given: A’=A ..(1)

⇒ (A²)’=(AA)’=A’A’

⇒ A’A’=(A)(A)=A²

⇒ (A²)’=A² (symmetric matrix) ..(2)

⇒ (A³)’=(A(A²))’=(A²)’A’

⇒ (A²)’A’=A²A= A³ (Using (1) and (2) )

⇒ (A³)’=A³ (symmetric matrix)

A³ is Also a symmetric matrix.

Given: A’=A ..(1)

⇒ (A²)’=(AA)’=A’A’

⇒ A’A’=(A)(A)=A²

⇒ (A²)’=A² (symmetric matrix) ..(2)

⇒ (A³)’=(A(A²))’=(A²)’A’

⇒ (A²)’A’=A²A= A³ (Using (1) and (2) )

⇒ (A³)’=A³ (symmetric matrix)

A² is a symmetric matrix.

Given: A’=-A

⇒ (A²)’=(AA)’=A’A’

⇒ A’A’=(-A)(-A)=A²

⇒ (A²)’=A² (symmetric matrix)

A² is a symmetric matrix.

Given: A’=-A

⇒ (A²)’=(AA)’=A’A’

⇒ A’A’=(-A)(-A)=A²

⇒ (A²)’=A² (symmetric matrix)

(i) (AB)’ = ________.

(AB)’ = B’A’

Let A be matrix of order m× n and B be of n× p.

A^’ is of order n× m and B^’ is of order p× n.

Hence B^’ A^’ is of order p× m.

So, AB is of order m× p.

And (AB)^’ is of order p× m.

We can see (AB)^’ and B^’ A^’ are of same order p× m.

Hence (AB)^’ = B^’ A^’

Hence proved.

(ii) (kA)’ = ________. (k is any scalar)

If a scalar “k” is multiplied to any matrix the new matrix becomes

K times of the old matrix.

Eg: A =

2A =

=

(2A)^’ =

A^’ =

Now 2A^’ =

=

Hence (2A)^’ =2A^’

Hence (kA)’ = k(A)’

(iii) [k (A – B)]’ = ________.

A =

A^’ =

2A^’ = 2

=

B=

B^’ =

2B^’ =

=

A-B =

Now Let k =2

2(A-B) =

=

[2(A-B)]^’ =

2A^’ – 2B^’ =

=

A^’ – B^’ =

=

2(A^’ – B^’) = 2

=

Hence we can see [k (A – B)]’= k(A)’- k(B)’= k(A’-B’)

(i) (AB)’ = ________.

(AB)’ = B’A’

Let A be matrix of order m× n and B be of n× p.

A^’ is of order n× m and B^’ is of order p× n.

Hence B^’ A^’ is of order p× m.

So, AB is of order m× p.

And (AB)^’ is of order p× m.

We can see (AB)^’ and B^’ A^’ are of same order p× m.

Hence (AB)^’ = B^’ A^’

Hence proved.

(ii) (kA)’ = ________. (k is any scalar)

If a scalar “k” is multiplied to any matrix the new matrix becomes

K times of the old matrix.

Eg: A =

2A =

=

(2A)^’ =

A^’ =

Now 2A^’ =

=

Hence (2A)^’ =2A^’

Hence (kA)’ = k(A)’

(iii) [k (A – B)]’ = ________.

A =

A^’ =

2A^’ = 2

=

B=

B^’ =

2B^’ =

=

A-B =

Now Let k =2